Question

Consider the cell reaction Zn(s)+Fe2+ (0.005M)↔Zn2+(0.01M) +Fe(s). Given

that the standard emf of the cell at 298K is 0.323 v, i) construct the cell ii)

Calculate emf of the cell.

Answer 1

i) Construct of the cell is:

ii) emf of the cell = standard emf + (0.059/n)*log(C_Fe2+/C_Zn2+)

standard emf = 0.323 V, n = 2 electrons, C_Fe2+ = 0.005 M, C_Zn2+ = 0.01 M, so:

emf of the cell = 0.323 + (0.059/2)*log(0.005/0.01) = 0.323 - 0.00888 = 0.314 V